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How do I go about simplifying this:

$$\sum_{k=1}^nk \cdot k!$$

Wolfram alpha tells me it's the same as $(n+1)!-1$ but I don't see how.

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  • $\begingroup$ If true it must follow by induction. $\endgroup$
    – David C. Ullrich
    Commented Dec 1, 2017 at 16:18

1 Answer 1

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Hint: $$k\cdot k! = [(k+1)-1]\cdot k! = (k+1)!-k!$$

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